Question

a study investigated about 300 meals ordered from chipotle restaurants using the online site grubhub. researchers calculated the sodium content (in milligrams) for each order based on chipotles published nutrition information. the distribution of sodium content is approximately normal with mean 2000 mg and standard deviation 500 mg.
(a) what proportion of the meals ordered exceeded the recommended daily allowance of 2400 mg of sodium?
(round to 4 decimal places.)
(b) what percent of the meals ordered had between 1200 mg and 1800 mg of sodium?
% (round to 4 decimal places then convert to a percentage.)

Explanation:

Step1: Calculate z - score for part (a)

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu = 2000$ mg (mean), $\sigma = 500$ mg (standard deviation), and $x = 2400$ mg. So, $z=\frac{2400 - 2000}{500}=\frac{400}{500}=0.8$.

Step2: Find the proportion for part (a)

We want $P(X>2400)$, which is equivalent to $1 - P(X\leq2400)$. Looking up the z - score of 0.8 in the standard normal table, $P(Z\leq0.8)=0.7881$. So, $P(X > 2400)=1 - 0.7881 = 0.2119$.

Step3: Calculate z - scores for part (b)

For $x_1 = 1200$ mg, $z_1=\frac{1200 - 2000}{500}=\frac{- 800}{500}=-1.6$. For $x_2 = 1800$ mg, $z_2=\frac{1800 - 2000}{500}=\frac{-200}{500}=-0.4$.

Step4: Find the proportion for part (b)

We want $P(1200

Answer:

(a) 0.2119
(b) 28.9800%